If you do not want to use "components", as Doc Al is suggesting (and using components is simpler), you can use the "sine law" and "cosine law" on the triangles formed.

In the first exercise you have a vector, u, that goes due east with "length" 10 and another vector, v, that goes 40 degrees W of N with "length" 15. If you draw those two sides of a triangle, the difference, v- u, is the third side of the triangle (directed from the tip of v to the tip of u). The length of that third side is given by the cosine law: [itex]c^2= a^2+ b^2- 2ab \cos(C)[/itex] where a and b are the two given lengths and C is the angle opposite side c. Here, the angle between the two given angle is 90+ 40= 130 degrees so [itex]c^2= 10^2+ 15^2- 2(10)(15)cos(130)[/itex]. You can then use the sine law to find the other two angles in the triangle and so deduce the direction of that vector.